Solve x^2(6%)^2+(1-x)^2(2%)^2+2x(1-x)*012*6%*2%=00327

$$x ^ { 2 } ( 6 \% ) ^ { 2 } + ( 1 - x ) ^ { 2 } ( 2 \% ) ^ { 2 } + 2 x ( 1 - x ) \times 012 \times 6 \% \times 2 \% = 00327$$

Solve for x (complex solution)

$$x=\frac{1}{10}+\frac{3}{10}i=0.1+0.3i$$

$$x=\frac{1}{10}-\frac{3}{10}i=0.1-0.3i$$

Steps Using the Quadratic Formula

Reduce the fraction \frac{6}{100} to lowest terms by extracting and canceling out 2.

$$x^{2}\times \left(\frac{3}{50}\right)^{2}+\left(1-x\right)^{2}\times \left(\frac{2}{100}\right)^{2}+2x\left(1-x\right)\times 0\times 12\times \frac{6}{100}\times \frac{2}{100}=0\times 0\times 327$$

Calculate \frac{3}{50} to the power of 2 and get \frac{9}{2500}.

$$x^{2}\times \frac{9}{2500}+\left(1-x\right)^{2}\times \left(\frac{2}{100}\right)^{2}+2x\left(1-x\right)\times 0\times 12\times \frac{6}{100}\times \frac{2}{100}=0\times 0\times 327$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-x\right)^{2}.

$$x^{2}\times \frac{9}{2500}+\left(1-2x+x^{2}\right)\times \left(\frac{2}{100}\right)^{2}+2x\left(1-x\right)\times 0\times 12\times \frac{6}{100}\times \frac{2}{100}=0\times 0\times 327$$

Reduce the fraction \frac{2}{100} to lowest terms by extracting and canceling out 2.

$$x^{2}\times \frac{9}{2500}+\left(1-2x+x^{2}\right)\times \left(\frac{1}{50}\right)^{2}+2x\left(1-x\right)\times 0\times 12\times \frac{6}{100}\times \frac{2}{100}=0\times 0\times 327$$

Calculate \frac{1}{50} to the power of 2 and get \frac{1}{2500}.

$$x^{2}\times \frac{9}{2500}+\left(1-2x+x^{2}\right)\times \frac{1}{2500}+2x\left(1-x\right)\times 0\times 12\times \frac{6}{100}\times \frac{2}{100}=0\times 0\times 327$$

Use the distributive property to multiply 1-2x+x^{2} by \frac{1}{2500}.

$$x^{2}\times \frac{9}{2500}+\frac{1}{2500}-\frac{1}{1250}x+\frac{1}{2500}x^{2}+2x\left(1-x\right)\times 0\times 12\times \frac{6}{100}\times \frac{2}{100}=0\times 0\times 327$$

Combine x^{2}\times \frac{9}{2500} and \frac{1}{2500}x^{2} to get \frac{1}{250}x^{2}.

$$\frac{1}{250}x^{2}+\frac{1}{2500}-\frac{1}{1250}x+2x\left(1-x\right)\times 0\times 12\times \frac{6}{100}\times \frac{2}{100}=0\times 0\times 327$$

Multiply 2 and 0 to get 0.

$$\frac{1}{250}x^{2}+\frac{1}{2500}-\frac{1}{1250}x+0x\left(1-x\right)\times 12\times \frac{6}{100}\times \frac{2}{100}=0\times 0\times 327$$

Multiply 0 and 12 to get 0.

$$\frac{1}{250}x^{2}+\frac{1}{2500}-\frac{1}{1250}x+0x\left(1-x\right)\times \frac{6}{100}\times \frac{2}{100}=0\times 0\times 327$$

Reduce the fraction \frac{6}{100} to lowest terms by extracting and canceling out 2.

$$\frac{1}{250}x^{2}+\frac{1}{2500}-\frac{1}{1250}x+0x\left(1-x\right)\times \frac{3}{50}\times \frac{2}{100}=0\times 0\times 327$$

Multiply 0 and \frac{3}{50} to get 0.

$$\frac{1}{250}x^{2}+\frac{1}{2500}-\frac{1}{1250}x+0x\left(1-x\right)\times \frac{2}{100}=0\times 0\times 327$$

Reduce the fraction \frac{2}{100} to lowest terms by extracting and canceling out 2.

$$\frac{1}{250}x^{2}+\frac{1}{2500}-\frac{1}{1250}x+0x\left(1-x\right)\times \frac{1}{50}=0\times 0\times 327$$

Multiply 0 and \frac{1}{50} to get 0.

$$\frac{1}{250}x^{2}+\frac{1}{2500}-\frac{1}{1250}x+0x\left(1-x\right)=0\times 0\times 327$$

Anything times zero gives zero.

$$\frac{1}{250}x^{2}+\frac{1}{2500}-\frac{1}{1250}x+0=0\times 0\times 327$$

Add \frac{1}{2500} and 0 to get \frac{1}{2500}.

$$\frac{1}{250}x^{2}+\frac{1}{2500}-\frac{1}{1250}x=0\times 0\times 327$$

Multiply 0 and 0 to get 0.

$$\frac{1}{250}x^{2}+\frac{1}{2500}-\frac{1}{1250}x=0\times 327$$

Multiply 0 and 327 to get 0.

$$\frac{1}{250}x^{2}+\frac{1}{2500}-\frac{1}{1250}x=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$\frac{1}{250}x^{2}-\frac{1}{1250}x+\frac{1}{2500}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{250} for a, -\frac{1}{1250} for b, and \frac{1}{2500} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-\frac{1}{1250}\right)±\sqrt{\left(-\frac{1}{1250}\right)^{2}-4\times \frac{1}{250}\times \frac{1}{2500}}}{2\times \frac{1}{250}}$$

Square -\frac{1}{1250} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\left(-\frac{1}{1250}\right)±\sqrt{\frac{1}{1562500}-4\times \frac{1}{250}\times \frac{1}{2500}}}{2\times \frac{1}{250}}$$

Multiply -4 times \frac{1}{250}.

$$x=\frac{-\left(-\frac{1}{1250}\right)±\sqrt{\frac{1}{1562500}-\frac{2}{125}\times \frac{1}{2500}}}{2\times \frac{1}{250}}$$

Multiply -\frac{2}{125} times \frac{1}{2500} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-\frac{1}{1250}\right)±\sqrt{\frac{1}{1562500}-\frac{1}{156250}}}{2\times \frac{1}{250}}$$

Add \frac{1}{1562500} to -\frac{1}{156250} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-\frac{1}{1250}\right)±\sqrt{-\frac{9}{1562500}}}{2\times \frac{1}{250}}$$

Take the square root of -\frac{9}{1562500}.

$$x=\frac{-\left(-\frac{1}{1250}\right)±\frac{3}{1250}i}{2\times \frac{1}{250}}$$

The opposite of -\frac{1}{1250} is \frac{1}{1250}.

$$x=\frac{\frac{1}{1250}±\frac{3}{1250}i}{2\times \frac{1}{250}}$$

Multiply 2 times \frac{1}{250}.

$$x=\frac{\frac{1}{1250}±\frac{3}{1250}i}{\frac{1}{125}}$$

Now solve the equation x=\frac{\frac{1}{1250}±\frac{3}{1250}i}{\frac{1}{125}} when ± is plus. Add \frac{1}{1250} to \frac{3}{1250}i.

$$x=\frac{\frac{1}{1250}+\frac{3}{1250}i}{\frac{1}{125}}$$

Divide \frac{1}{1250}+\frac{3}{1250}i by \frac{1}{125} by multiplying \frac{1}{1250}+\frac{3}{1250}i by the reciprocal of \frac{1}{125}.

$$x=\frac{1}{10}+\frac{3}{10}i$$

Now solve the equation x=\frac{\frac{1}{1250}±\frac{3}{1250}i}{\frac{1}{125}} when ± is minus. Subtract \frac{3}{1250}i from \frac{1}{1250}.

$$x=\frac{\frac{1}{1250}-\frac{3}{1250}i}{\frac{1}{125}}$$

Divide \frac{1}{1250}-\frac{3}{1250}i by \frac{1}{125} by multiplying \frac{1}{1250}-\frac{3}{1250}i by the reciprocal of \frac{1}{125}.

$$x=\frac{1}{10}-\frac{3}{10}i$$

The equation is now solved.

$$x=\frac{1}{10}+\frac{3}{10}i x=\frac{1}{10}-\frac{3}{10}i$$

Steps for Completing the Square

Reduce the fraction \frac{6}{100} to lowest terms by extracting and canceling out 2.

$$x^{2}\times \left(\frac{3}{50}\right)^{2}+\left(1-x\right)^{2}\times \left(\frac{2}{100}\right)^{2}+2x\left(1-x\right)\times 0\times 12\times \frac{6}{100}\times \frac{2}{100}=0\times 0\times 327$$

Calculate \frac{3}{50} to the power of 2 and get \frac{9}{2500}.

$$x^{2}\times \frac{9}{2500}+\left(1-x\right)^{2}\times \left(\frac{2}{100}\right)^{2}+2x\left(1-x\right)\times 0\times 12\times \frac{6}{100}\times \frac{2}{100}=0\times 0\times 327$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-x\right)^{2}.

$$x^{2}\times \frac{9}{2500}+\left(1-2x+x^{2}\right)\times \left(\frac{2}{100}\right)^{2}+2x\left(1-x\right)\times 0\times 12\times \frac{6}{100}\times \frac{2}{100}=0\times 0\times 327$$

Reduce the fraction \frac{2}{100} to lowest terms by extracting and canceling out 2.

$$x^{2}\times \frac{9}{2500}+\left(1-2x+x^{2}\right)\times \left(\frac{1}{50}\right)^{2}+2x\left(1-x\right)\times 0\times 12\times \frac{6}{100}\times \frac{2}{100}=0\times 0\times 327$$

Calculate \frac{1}{50} to the power of 2 and get \frac{1}{2500}.

$$x^{2}\times \frac{9}{2500}+\left(1-2x+x^{2}\right)\times \frac{1}{2500}+2x\left(1-x\right)\times 0\times 12\times \frac{6}{100}\times \frac{2}{100}=0\times 0\times 327$$

Use the distributive property to multiply 1-2x+x^{2} by \frac{1}{2500}.

$$x^{2}\times \frac{9}{2500}+\frac{1}{2500}-\frac{1}{1250}x+\frac{1}{2500}x^{2}+2x\left(1-x\right)\times 0\times 12\times \frac{6}{100}\times \frac{2}{100}=0\times 0\times 327$$

Combine x^{2}\times \frac{9}{2500} and \frac{1}{2500}x^{2} to get \frac{1}{250}x^{2}.

$$\frac{1}{250}x^{2}+\frac{1}{2500}-\frac{1}{1250}x+2x\left(1-x\right)\times 0\times 12\times \frac{6}{100}\times \frac{2}{100}=0\times 0\times 327$$

Multiply 2 and 0 to get 0.

$$\frac{1}{250}x^{2}+\frac{1}{2500}-\frac{1}{1250}x+0x\left(1-x\right)\times 12\times \frac{6}{100}\times \frac{2}{100}=0\times 0\times 327$$

Multiply 0 and 12 to get 0.

$$\frac{1}{250}x^{2}+\frac{1}{2500}-\frac{1}{1250}x+0x\left(1-x\right)\times \frac{6}{100}\times \frac{2}{100}=0\times 0\times 327$$

Reduce the fraction \frac{6}{100} to lowest terms by extracting and canceling out 2.

$$\frac{1}{250}x^{2}+\frac{1}{2500}-\frac{1}{1250}x+0x\left(1-x\right)\times \frac{3}{50}\times \frac{2}{100}=0\times 0\times 327$$

Multiply 0 and \frac{3}{50} to get 0.

$$\frac{1}{250}x^{2}+\frac{1}{2500}-\frac{1}{1250}x+0x\left(1-x\right)\times \frac{2}{100}=0\times 0\times 327$$

Reduce the fraction \frac{2}{100} to lowest terms by extracting and canceling out 2.

$$\frac{1}{250}x^{2}+\frac{1}{2500}-\frac{1}{1250}x+0x\left(1-x\right)\times \frac{1}{50}=0\times 0\times 327$$

Multiply 0 and \frac{1}{50} to get 0.

$$\frac{1}{250}x^{2}+\frac{1}{2500}-\frac{1}{1250}x+0x\left(1-x\right)=0\times 0\times 327$$

Anything times zero gives zero.

$$\frac{1}{250}x^{2}+\frac{1}{2500}-\frac{1}{1250}x+0=0\times 0\times 327$$

Add \frac{1}{2500} and 0 to get \frac{1}{2500}.

$$\frac{1}{250}x^{2}+\frac{1}{2500}-\frac{1}{1250}x=0\times 0\times 327$$

Multiply 0 and 0 to get 0.

$$\frac{1}{250}x^{2}+\frac{1}{2500}-\frac{1}{1250}x=0\times 327$$

Multiply 0 and 327 to get 0.

$$\frac{1}{250}x^{2}+\frac{1}{2500}-\frac{1}{1250}x=0$$

Subtract \frac{1}{2500} from both sides. Anything subtracted from zero gives its negation.

$$\frac{1}{250}x^{2}-\frac{1}{1250}x=-\frac{1}{2500}$$

Multiply both sides by 250.

$$\frac{\frac{1}{250}x^{2}-\frac{1}{1250}x}{\frac{1}{250}}=-\frac{\frac{1}{2500}}{\frac{1}{250}}$$

Dividing by \frac{1}{250} undoes the multiplication by \frac{1}{250}.

$$x^{2}+\left(-\frac{\frac{1}{1250}}{\frac{1}{250}}\right)x=-\frac{\frac{1}{2500}}{\frac{1}{250}}$$

Divide -\frac{1}{1250} by \frac{1}{250} by multiplying -\frac{1}{1250} by the reciprocal of \frac{1}{250}.

$$x^{2}-\frac{1}{5}x=-\frac{\frac{1}{2500}}{\frac{1}{250}}$$

Divide -\frac{1}{2500} by \frac{1}{250} by multiplying -\frac{1}{2500} by the reciprocal of \frac{1}{250}.

$$x^{2}-\frac{1}{5}x=-\frac{1}{10}$$

Divide -\frac{1}{5}, the coefficient of the x term, by 2 to get -\frac{1}{10}. Then add the square of -\frac{1}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{5}x+\left(-\frac{1}{10}\right)^{2}=-\frac{1}{10}+\left(-\frac{1}{10}\right)^{2}$$

Square -\frac{1}{10} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{5}x+\frac{1}{100}=-\frac{1}{10}+\frac{1}{100}$$

Add -\frac{1}{10} to \frac{1}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{5}x+\frac{1}{100}=-\frac{9}{100}$$

Factor x^{2}-\frac{1}{5}x+\frac{1}{100}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{10}\right)^{2}=-\frac{9}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{10}\right)^{2}}=\sqrt{-\frac{9}{100}}$$

Simplify.

$$x-\frac{1}{10}=\frac{3}{10}i x-\frac{1}{10}=-\frac{3}{10}i$$

Add \frac{1}{10} to both sides of the equation.

$$x=\frac{1}{10}+\frac{3}{10}i x=\frac{1}{10}-\frac{3}{10}i$$